Resource Allocation Merton Case_Group 5.pptx

Merton Case

1. Merton Truck Company Case EXERCISE Group 5 R. Sukma Nugraha A – 15122320025 Reynard Nathaniel – 15122320027 Sarah Hanifa Purnomo – 15122320045 Tri Wicaksono – 15122320041

2. Merton Truck Company 2 Background Case : Merton truck company was unable to find the best product mix for their model : 101 and 102, the point where they can maximize their profits while keeping optimal capacity utilization for the four manufacturing departments including engine assembly, metal stamping, Model 102 assembly, and Model 101 assembly. Available Information : Machine hour per truck requirement & maximum capacity for each manufacturing processes: Financial statements show that Model 101 truck has a lower contribution margin than Model 102 truck. (Depicted unit price, standard product cost & overhead budget ) Fixed Cost Overhead $8,600,000 Description Machine hr required per truck (101) Machine hr required per truck (102) Maximum Capacity  Engine assembly                               1                             2                       4,000  Metal stamping                               2                             2                       6,000  Model 101 assembly                               2                            -                       5,000  Model 102 assembly                              -                             3                       4,500

3. 1a. Find the best product mix for Merton? 3 Data tables

4. 1a. Find the best product mix for Merton Tambahkan footer 4 Maximize Z = 3000X1 + 3=5000X2 Subject to: X1+2X2<=4000 2X1+2X2<=6000 2X1 <= 4000 3X2 <= 4500 Where: X1 = Model 101 X2 = Model 102 Maximize Z = 2000*3000 + 1000*5000 = 11,000,000 (before fixed overhead) Conclusion: In order to maximize cost, the mix product consist of : 2000 Model 101 1000 Model 102 The profit is $11,000,000 - &86,000,000 =$ 2 ,400,000

5. 1b. Find the best product mix if engine assembly capacity is raised to 4001 machine-hours Tambahkan footer 5 Conclusion: In order to maximize cost, the mix product consist of : 1999 Model 101 1001 Model 102 The profit is $11,002,000 - $8,600.000 = $2,402,000 An increase of $2000 per increase per 1 machine-hour Maximize Z = 3000X1 + 3=5000X2 Subject to: X1+2X2<=4001 2X1+2X2<=6000 2X1 <= 4000 3X2 <= 4500 Where: X1 = Model 101 X2 = Model 102 Maximize Z = 2000*3000 + 1000*5000 = 11,002,000 (before fixed overhead)

6. 1c. If the engine capacity is increased to 4100 hours, verify the contribution will also increase 100x Tambahkan footer 6 Conclusion: The profit is $11,200,000 - $8,600.000 = $2,600,000 An increase of $200,000 per increase per 100 machine-hours An equivalent of of $2000 per increase per 1 machine-hour from result 1b It is proven a 100x increase in machine hours is equivalent of 100x increase in contribution. Maximize Z = 3000X1 + 3=5000X2 Subject to: X1+2X2<=4100 2X1+2X2<=6000 2X1 <= 4000 3X2 <= 4500 Where: X1 = Model 101 X2 = Model 102 Maximize Z = 2000*3000 + 1000*5000 = 11,200,000 (before fixed overhead)

7. 1d. How much can be added into engine assembly capacity before there is a change in value per unit of capacity? Tambahkan footer 7 Conclusion: Engine assembly hours has an allowable increase of 500 hours , and an upper bound of 4500 hours before there is a change in value per machining hour

8. 1d. How much can be added into engine assembly capacity before there is a change in value per unit of capacity? Tambahkan footer 8 Conclusion: A machine assembly capacity of 4501 will result in the same profit as a capacity of 4500, but will result in a surplus in machining hours. Going over 4500 hours has resulted in wasted resources

9. Merton Truck Company - Case Number 2 9 Conclusion: The company should adopt this alternative (consider renting) since it rises $1,000,000 profit. The maximum rent it should be willing to pay for a machine-hour of engine assembly capacity is $2,000. The maximum number of machine-hours it should rent is 500/hr . Assumptions: If Merton's decides to increase the capacity the engine assembly department to maximum capacity Maximize profit Z = 3000X1 + 5000X2 Subject to: X1 + 2X2 ≤ 4500 (Engine Assembly) 2X1 + 2X2 ≤ 6000 (Metal Stamping) 2X1 ≤ 5000 (101 Assembly) 3X2 ≤ 4500 (102 Assembly) X1, X2 ≥ 0 (Non negativity)

10. Merton Truck Company - Case Number 3a 10 Conclusion: Merton should not produce Model 103 trucks since the contribution is too small to increase the profit. Assumptions: Merton's considering new model, 103. And Model 103 would give contribution $2000 Maximize profit Z = 3000X1 + 5000X2 + 2000X3 Subject to: X1 + 2X2 + 0,8X3 ≤ 4000 (Engine Assembly) 2X1 + 2X2 + 1,5X3 ≤ 6000 (Metal Stamping) 2X1+ 1X3 ≤ 5000 (101 Assembly) 3X2 ≤ 4500 (102 Assembly) X1, X2, X3 ≥ 0 (Non negativity)

11. Merton Truck Company - Case Number 3b 11 Conclusion: The contribution on each Model 103 truck have to be $2,350 before it become worthwhile to produce the new model. If the contribution on each Model 103 truck is $2,351, the production immediately changed into 2,857 Model 103 trucks and 857 Model 102 trucks with $2,402,857 profit ($2,857 more). It may occurred because the machine-hours of engine assembly, metal stamping, and Model 103 assembly is more efficient. Assumptions: How high would the contribution of Model 103 have to be before it became wothwile to produce the new model? Maximize profit Z = 3000X1 + 5000X2 + 2350X3 Subject to: X1 + 2X2 + 0,8X3 ≤ 4000 (Engine Assembly) 2X1 + 2X2 + 1,5X3 ≤ 6000 (Metal Stamping) 2X1+ 1X3 ≤ 5000 (101 Assembly) 3X2 ≤ 4500 (102 Assembly) X1, X2, X3 ≥ 0 (Non negativity)

12. Merton Truck Company – SR Case Number 3 12

13. Merton Truck Company - Case Number 4 13 A ssumptions : 1. P roduction cost efficiency does not change , 2. 2,000 machine hours of engine assembly overtime are available 3. D irect labor costs increase 50% 4.Monthly f ixed overhead costs increase $750,000 Maximize Z = 3.000X1 + 5000X2 + 2.400X3 + 3.800X4 Subject to: X1 + 2X2 ≤ 4.000 (Engine AssemblY X3 + 2X4 ≤ 2,000 (Engine Assembly Overtime) 2X1+ 2X2 + 2X3 + 2X4 ≤ 6,000 (Metal Stamping) 2X2 + 2X3 ≤ 5,000 (Model 101Assembling) 3X2 +3X4 ≤ 4,500 (Model 102 Assembling) X1, X2, X3, X4 ≥ 0 (Non negativity) Maximize Z = 3.000*1.500 + 5.000*1.250 + 2.400*0 + 3.800*250 = 11.700.000 Conclusion: Based on the simulation using the assumptions , Merton is not recommended to assemble the model 101 and model 102 using overtime because it will increase fixed overhead $750,000 which is higher than profits $700,000 . .

14. Merton Truck Company 14

15. 5. Management proposal to produce 101 Models at least 3 times the number of Model 102. Tambahkan footer 15 Maximize Z = 3000X1 + 3=5000X2 Subject to: X1+2X2<=4000 2X1+2X2<=6000 2X1 <= 4000 3X2 <= 4500 X1-3X2<=0 Maximize Z = 2250 *3000 + 750*5000 = 1 0,500,000 Conclusion: The proposal to produce 101 model 3 times of model 102 is yielding profit $10,500,000 which is lower than the optimum profit that has been generated in case 1 (2000 Model 101 & 1000 Model 102 yielding profit $11,000,000 ). Therefore the proposal is not feasible to be executed.

16. Thank You 16